Problems and Solutions in - Mathematics - University of Hawaii

1.1 1991 November 21 1 REAL ANALYSIS
Let n > 0 be such that
X\An
|f| dm < ɛ/2.
Define δ = (2n) −1 ɛ, and suppose E ⊂ [0, 1] is a measurable set with mE < δ. We must show |
E
f dm
≤
=
≤
E
|f| dm
(X\An)∩E
X\An
|f| dm +
|f| dm +
< ɛ/2 + n m(An ∩ E)
< ɛ/2 + n ɛ
= ɛ.
2n
An∩E
An∩E
|f| dm
|f| dm
E
f dm| < ɛ.
The penultimate inequality holds because |f| < n on An. ⊓⊔
Proof 2: 7 This proof relies on the following lemma about absolute continuity of measures:
Lemma 1.1 Let ν be a finite signed measure and µ a positive measure on a measurable space (X, M). Then ν ≪ µ
if and only if for every ɛ > 0 there is a δ > 0 such that |ν(E)| < ɛ whenever µ(E) < δ.
The signed measure defined by
ν(E) =
is finite iff 8 f ∈ L 1 (µ). It is also clearly absolutely continuous with respect to µ. Therefore, lemma 1.1 can be
applied to the real and imaginary parts of any complex-valued f ∈ L1 (µ). It follows that, for every ɛ > 0, there is
a δ > 0 such that
|ν(E)| =
f dµ
< ɛ, whenever µ(E) < δ.
E
7. 9 Suppose f is a bounded, real valued, measurable function on [0, 1] such that x n f dm = 0 for n = 0, 1, 2, . . .,
with m Lebesgue measure. Show that f(x) = 0 a.e.
Solution: Fix an arbitrary continuous function on [0, 1], say, φ ∈ C[0, 1]. By the Stone-Weierstrass theorem, there
is a sequence {pn} of polynomials such that φ − pn∞ → 0 as n → ∞. Then, since all functions involved are
7 See also Folland [4], page 89.
8 For what follows we only need that ν is finite if f is integrable, but the converse is also true.
9 This question appears very often in varying forms of difficulty. cf. November ’92 (7b, very easy version), November ’96 (B2, very easy),
November ’91 (this question, easy), April ’92 (6, moderate), November ’95 (6, hard–impossible?). I have yet to solve the November ’95 version.
One attempted solution (which I think is the one given in the black notebook), seems to assume f ∈ L 1 , but that assumption makes the problem
even easier than the others.
6
E
f dµ
⊓⊔